This is a question from CAT Mensuration. In this question, a ball is kept on the top of a hollow cylinder. We need to find the distance, of the topmost point of the ball from the base of the cylinder. Mensuration Questions in the CAT exam can be solved with a practical approach. Even if one is not completely prepared for CAT Geometry, he or she should be able to do well in Mensuration problems tested by the **CAT Exam**. This is one of the questions which looks difficult on first look but with a little visualization and concept, can be cracked in a minute.

Question 19 : A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 π cm^{3}. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is: [TITA]

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Given that the height of the cylinder is 3 cm, while its volume is 9π cm^{3}

Volume of the cylinder ⟹ πr^{2}h = 9π cm^{3}

r^{2} = 9 × 3 or r = √3

So the diameter is 2√3 cm

A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically.

Since OPQ is the right angled triangle

We can find the OP = 1 cm.

We have to find the vertical distance of the topmost point of the ball from the base of the cylinder.

Since OP = 1, to reach the topmost point still it has to go 2 cm from the point O.

The vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is 2 + 1 + 3 = 6 cm

The question is **"A ball of diameter 4 cm is kept on top of a hollow cylinder standing vertically. The height of the cylinder is 3 cm, while its volume is 9 π cm ^{3}. Then the vertical distance, in cm, of the topmost point of the ball from the base of the cylinder is:" **

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